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mat25-hw6solns - Math 25 Advanced Calculus UC Davis Spring...

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Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 – Solutions to Homework Assignment #6 1. Let sequences ( s n ) n =1 and ( t n ) n =1 converge to respective limits S and T . Fix some > 0. By definition, there exist numbers m s , m t such that for any n s m s and n t m t , | s n s - S | < 2 and | t n t - T | < 2 . If n max { m s , m t } , by the triangle inequality, | ( s n - t n ) - ( S - T ) | = | ( s n - S ) - ( t n - T ) | ≤ | s n - S | + | t n - T | < 2 + 2 = . By definition, then, s n - t n -→ S - T as n -→ ∞ . 2. For ( s n ) n =1 and ( t n ) n =1 such that s n = n + 1 , t n = n, Both sequences diverge to infinity, and hence the reasoning in problem 1. does not apply. However, for any n , since a - b = ( a - b )( a + b ), s n - t n = n + 1 - n = ( n + 1) - n n + 1 + n = 1 n + 1 + n Now, since 0 1 n + 1 + n 1 n And because 1 n -→ 0 as n -→ ∞ , by the squeeze theorem, ( s n - t n ) n =1 converges to 0. 3. (a) lim n →∞ 1 n 3 = 0, since 1 n 3 1 n , and 1 n -→ 0 as n -→ ∞ .
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